InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1551. |
In sugarcane plant `^(14)CO_(2)` is fixed in malic acid, in which the enzyme that fixes `CO_(2)` isA. Ribulose biphosphate carboxylaseB. Phosphoenol pyruvic acid carboxylaseC. Ribulose phosphate kinaseD. Fructose phosphatase |
| Answer» Correct Answer - B | |
| 1552. |
In the leaves of `C_(4)` plants, malic acid formation during `CO_(2)` fixation occurs in the cells ofA. bundle sheathB. guard cellsC. epidermal cellsD. mesophyll cells |
| Answer» Correct Answer - D | |
| 1553. |
Which one of the following compounds is not a protoric acid?A. `B(OH)_3`B. `PO(OH)_3`C. `SO(OH)_2`D. `SO_2(OH)_2` |
| Answer» Correct Answer - A | |
| 1554. |
Which one of the following characteristics of the transition metals is associated with their catalytic activity?A. High enthalpy of atomizationB. Paramagnetic behaviourC. Colour of hydrated ionsD. Variable oxidation states |
| Answer» Correct Answer - D | |
| 1555. |
The basic character of the transition metal monoxide follows the orderA. VO gt CrO gt TiO gt FeOB. CrO gt VO gt FeO gt TiOC. TiO gt FeO gt VO gt CrOD. TiO gt VO gt CrO gt FeO |
| Answer» Correct Answer - D | |
| 1556. |
How many grams of concentrated nitric acid solution should be used to prepare `250 mL` of `2.0 M HNO_(3)`? The concentrated acid is `70% HNO_(3)`:A. `45.0" g conc. HNO"_(3)`B. `90.0" g conc. HNO"_(3)`C. `70.0" conc. HNO"_(3)`D. `54.0" g conc. HNO"_(3)` |
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Answer» Correct Answer - A `(w)/(E)=NV` `(w)/(6.3)=2xx(250)/(1000)` `w=(63)/(2)=31.5g` but `70%` solution 70 g is used the wt. of solution is 100 g 31.5 g is used the wt. of solution is ? `(100)/(70)xx31.5=45g` |
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| 1557. |
`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:A. 0.955 M and 1.910 MB. 1.910 M and 0.955 MC. 1.90 M and 1.910 MD. 0.477 and 0.477 M |
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Answer» Correct Answer - B Molarity `(M )=(wt)/(mol .wt.) (1000)/(vol (ml))` `=(25.3)/(106)xx(1000)/(250)` `=.955mol//L`of `Na_(2)CO_(3)` and `Na_(2)CO_(3) to 2Na^(+)+CO_(3)^(-2)` therefore`[Na^(+)]=2xx0.955=1.910M` `[CO_(3)^(-2)]=0.955M` |
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| 1558. |
When dominant and recessive alleles express themselves together, it is calledA. Co-dominanceB. DominanceC. AmphidominanceD. Pseudo dominance |
| Answer» Correct Answer - A | |
| 1559. |
Alzhimer disease in humans is associated with the deficiency ofA. Gamma aminobutyric acid (GABA)B. DopamineC. Glutamic acidD. Acetylcholine |
| Answer» Correct Answer - 4 | |
| 1560. |
The plasma membrane consists mainly ofA. phospholipids embedded in a protein bilayerB. proteins embedded in a phospholipid bilayerC. proteins embedded in a polymer of glucose moleculesD. proteins embedded in a carbohydrate bilayer |
| Answer» Correct Answer - B | |
| 1561. |
Which one of the following does not act as a neurotransmitter ?A. AcetylcholineB. EpinephrineC. NorepinephrineD. Cortisone |
| Answer» Correct Answer - D | |
| 1562. |
The Earth Summit held in Rio de Janeiro in 1992 was called :A. for immediate steps to discontinue use of CFCs that were damaging the ozone layer.B. to reduce `CO_(2)` emissions and global warmingC. for conservation of biodiversity and sustainable utilization of its benefits.D. to assess threat posed to native species by invasive weed species. |
| Answer» Correct Answer - C | |
| 1563. |
A coil of `40 H` inductance is connected in series with a resistance of `8` ohm and the combination is joined to the terminals of a `2 V` battery. The time constant of the circuitA. 1/5 secondsB. 40 secondsC. 20 secondsD. 5 seconds |
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Answer» Correct Answer - D Time constant `=L/R=40/8`=5 seconds |
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| 1564. |
An electric dipole has the magnitude of its charge as `q` and its dipole moment is `p`. It is placed in a uniform electric field `E`. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectivelyA. q. E and p. EB. zero and minimumC. q. E and maximumD. 2q. E and minimum |
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Answer» Correct Answer - B `F=p"dE"/"dr"=0` (`because` E= constant ) `u=-vecp.vecE`=-PE (minimum ) |
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| 1565. |
The magnetic flux through a circuit of resistance `R` changes by an amount `Deltaphi` in a time `Deltat`. Then the total quantity of electric charge `Q` that passes any point in the circuit during the time `Deltat` is represent byA. `Q=(Deltaphi)/R`B. `Q=(Deltaphi)/(Deltat)`C. `Q=R.(Deltaphi)/(Deltat) `D. `Q=1/R. (Deltaphi)/(Delta t)` |
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Answer» Correct Answer - A `I=e/R=I/R((Deltaphi)/(Deltat))=(DeltaQ)/(Deltat) rArr Q = (Deltaphi)/R` |
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| 1566. |
A bullet of mass `2 gm` is having a charge of `2 muc`. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of `10 m//s`A. 50 kVB. 5VC. 50 VD. 5kV |
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Answer» Correct Answer - A use `qV_"acc"=1/2mv^2` and get required result |
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| 1567. |
The electric field associted with an electromagnetic wave in vacuum is given by `vecE=hati 40 cos(kz=6xx10^(8)t)`, when `E`, `z` and `t` are in volt/m metre and second respectivelyfind the wave vector.A. `6m^(-1)`B. `3m^(-1)`C. `2 m^(-1)`D. `0.5 m^(-1)` |
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Answer» Correct Answer - C Equation of Electromagnetic wave is `E=E_(0) cos (kz-omega t)` and speed of EM wave `V=(omega)/(k)` By comparing `k=omega//v=(6xx10^(8))/(3xx10^(8))=2 m^(-1)` is the wave vector. |
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| 1568. |
For a cubic crystal structure which one of the following relations indicating the cell characterstic is correct?A. ` a=b=c and alpha = beta= gamma =90^(@) `B. `a ne b ne c and alpha ne beta ne gamma ne 90^(@) `C. `a ne be ne c and alpha = beta = gamma =90^(@)`D. `a=b=c and alpha ne beta ne gamma =90^(@) ` |
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Answer» Correct Answer - A For cubic crystal structure a=b=c= & `alpha=beta=gamma=90^(@)` |
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| 1569. |
A body of mass `1 kg` is thrown upwards with a velocity `20 ms^(-1)`. It momentarily comes to rest after attaining a height of `18 m`. How much energy is lost due to air friction? `(g = 10 ms^(-2))`A. 10JB. 20JC. 30JD. 40J |
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Answer» Correct Answer - B Loss of energy `=(1)/(2)mv^(2)-mgh=20J` |
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| 1570. |
The additional kinetic energy to be provided to a satellite of mass `m` revolving around a planet of mass `M`, to transfer it forms a circular orbit of radius `R_(1)` to another of radius `R_(2)(R_(2)gtR_(1))` isA. `GmM((1)/(R_(1))-(1)/(R_(2)))`B. `2 GmM((1)/(R_(1))-(1)/(R_(2)))`C. `(1)/(2)GmM((1)/(R_(1))-(1)/(R_(2)))`D. `GmM((1)/(R_(1)^(2))-(1)/(R_(2)^(2)))` |
| Answer» Correct Answer - C | |
| 1571. |
What is the direction of micropyle in anatropous ovuleA. UpwardB. DownwardC. RightD. Left |
| Answer» Correct Answer - B | |
| 1572. |
In the five - kingdom classification Chlamydomonas and Chlorella have been included inA. AlgaeB. PlantaeC. MoneraD. Protista |
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Answer» Correct Answer - D According to 5 kingdom classification Chlamydomonas and Chlorella are the members of Protista. |
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| 1573. |
Grey crescent is the areaA. At the point of entry of sperm into ovumB. Just opposite to the site of entry of sperm into ovumC. At the animal poleD. At the vegetal pole |
| Answer» Correct Answer - B | |
| 1574. |
For the synthesis of one glucose molecule the calvin cycle operates for:A. 2 timesB. 4 timesC. 6 timesD. 8 times |
| Answer» Correct Answer - C | |
| 1575. |
A sedentary sea anemone gets attached to the shell lining of hermit crab. The association isA. EctoparasitismB. SymbiosisC. CommensalismD. Amensalism |
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Answer» Correct Answer - B The association between sea anemone and hermit crab is symbiosis as both live together for very long duration |
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| 1576. |
When the plants are grown in magnesium deficient but urea rich soil, the symptoms expressed are:A. Yellowish leavesB. Colourless petioleC. Dark green leavesD. Shoot apex die |
| Answer» Correct Answer - A | |
| 1577. |
The first step for initiation of photsynthesis will be:A. Photolysis of waterB. Excitement of chloroBIOll molecule due to absorption of lightC. ATP formationD. Glucose formation |
| Answer» Correct Answer - B | |
| 1578. |
Artificial selection to obtain cows yielding higher milk output representsA. Stabilizing selection as it stabilizes this character in the populationB. Directional as it pushes the mean of the character in one directionC. Disruptive as it splits the population into two one yielding higher output and the other lower outputD. Stabilizing followed by disruptive as it stabilizes the population to produce higher yielding cows |
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Answer» Correct Answer - B Artificial selection to obtain cow yielding higher milk output will shift the peak to one direction, hence, will be an example of Directional selection. In stabilizing selection, the organisms with the mean value of the trait are selected. In disruptive selection, both extremes get selected. |
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| 1579. |
The total number of bones in your right arm is " " Or Total number of bones in the hind limb of a man isA. 14B. 24C. 26D. 30 |
| Answer» Correct Answer - D | |
| 1580. |
Which of the following stimulates the secretion of gastric juice :A. GastrinB. EnterogasteroneC. SecretinD. Hepatocrinin |
| Answer» Correct Answer - A | |
| 1581. |
Vitamin which induced maturation of R.B.C. `:-`A. `B_(1)`B. AC. `B_(12)`D. D |
| Answer» Correct Answer - C | |
| 1582. |
Which one of the following sequence was proposed by Darwin and Wallace for organic evolutionA. Overproduction, variations, constancy of population size, natural selectionB. Variations, constancy of population size, overproduction, natural selectionC. Overproduction, constancy of population size, variations, natural selectionD. Variations, natural selection, overproduction, constancy of population size |
| Answer» Correct Answer - C | |
| 1583. |
In stratosphere , which of the following element acts as a catalyst in degradation of ozone an release of molecular oxygen ?A. FeB. ClC. CarbonD. Oxygen |
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Answer» Correct Answer - B UV rays act on CFCs, releasing Cl atoms, chlorine reacts with ozone in sequential method converting into oxygen Carbon, oxygen and Fe are not related to ozone layer depletion |
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| 1584. |
In which one of the following the BOD (Biochemical Oxygen Demand) of sewage (S), distillery effluent (DE), paper mill effluent (PE) and sugar mill effluent (SE) have been arranged in ascending order ?A. `S lt DE lt PE lt SE`B. `SE lt S lt PE lt DE`C. `SE lt PE lt S lt DE`D. `PE lt S lt SE lt DE` |
| Answer» Correct Answer - B | |
| 1585. |
A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again:-A. 1 cm upwardB. 4.5 cm downwardC. 1 cm downwardD. 2 cm upward |
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Answer» Correct Answer - A Shifting in microscope=upward shifting in mark `=t(1-1/mu)=3(1-(1)/1.5)=1cm` |
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| 1586. |
`0.002m` aqueous solution of an ionic compound `Co(NH_(3))_(5)(NO_(2))CI` freezes at `-0.00732^(@)C`.Number of moles of ions which 1 mole of ionic compound produces in water will be `(K_(f) = 1.86^(@)C//m)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `DeltaT_(f)=ik_(f).m` `i=(DeltaT_(f))/(k_(f).m)` `=(0.00732)/(1.86xx0.002)=(0.00732)/(0.00372)` `i=2` Compound will be `[Co(NH_(3))_(5)]NO_(5)NO_(2)]Cl` Total possible ions =2 |
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| 1587. |
The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :A. `1.25xx 10^(-4)`B. `1.25xx10^(-5)`C. `1.25 xx 10^(-6)`D. `6.25xx10^(-4)` |
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Answer» Correct Answer - B `alpha=(A_(C))/(A_(oo))=(8)/(400)` `K_(a)=C alpha^(2)` `=(1)/(32)xx(8)/(400)xx(8)/(400)` `=1.25xx10^(-5)` |
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| 1588. |
The Lewis acid character of boron trihalides decreases as: `B Br_(3) gt BCl_(3) gt BF_(3)`. Explain ?A. ` BCl_(3) gt BF_(3) gt BBr_(3)`B. `BBr_(3) gt BCl_(3) gt BF_(3 )`C. ` BBr_(3) gt BF_(3) gt BCl_(3 )`D. |
| Answer» Correct Answer - B | |
| 1589. |
Which of the following has `ppi-dpi` bonding?A. `NO_3^-`B. `SO_3^(-2)`C. `BO_3^(-3)`D. `CO_3^(-2)` |
| Answer» Correct Answer - B | |
| 1590. |
Which one of the following arrangements does not give the correct picture of the trends indicated against it? (a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy (b) F2 > Cl2 > Br2 > I2 : Electronegativity (c) F2 > Cl2 > Br2 > I2 : Oxidizing power (d) F2 > Cl2 > Br2 > I2 : Eiectron gain enthalpy |
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Answer» The correct option is : (a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy (d) F2 > Cl2 > Br2 > I2 : Eiectron gain enthalpy Explanation: In case of diatomic molecules (X2) of halogens the bond dissociation energy decreases in the order : Cl2 > Br2 > F2 > I2 The oxidising power, electronegativity and reactivity decrease in the order : F2 > Cl2 > Br2 > I2 Electron gain enthalpy of halogens follows the given order: Cl2 > F2 > Br2 > I2 The low value of electron gain enthalpy (electron enthalpy) of fluorine is probably due to small size of fluorine atom. |
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| 1591. |
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O? (a) Al < Ca < O < C < F (b)Al < O < C < Ca < F (c) C < F < O < Al < Ca (d) Ca < Al < C < O < F |
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Answer» The correct option is : (d) Ca < Al < C < O < F Explanation: Electron gain enthalpy becomes less negative from top to bottom in a group while it becomes more negative from left to right within a period. |
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| 1592. |
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is(a) Mg < Ca < Cl < P (b) Cl < P < Mg < Ca(c) P < Cl < Ca < Mg (d) Ca < Mg < P < Cl |
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Answer» The correct option is : (b) Cl < P < Mg < Ca Explanation: The atomic radii decrease on moving from left to right in a period, thus order of sizes for Cl, P and Mg is Cl < P < Mg. Down the group size increases. Thus overall order is : Cl < P < Mg < Ca. |
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| 1593. |
The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase? (a) Ca2+ < K+ < Ar (b) K+ < Ar < Ca2+ (c) Ar < K+ < Ca2+ (d) Ca2+ < Ar < K+ |
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Answer» The correct option is : (a) Ca2+ < K+ < Ar Explanation : In case of isoelectronic species, radius decreases with increase in nuclear charge. |
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| 1594. |
In which of the following options the order of arrangement does not agree with the variation of property indicated against it?(a) I < Br < Cl < F (increasing electron gain enthalpy)(b) Li < Na < K < Rb (increasing metallic radius)(c) Al3+ = Mg2+ < Na+ < F- (increasing ionic size)(d) B < C < N < O (increasing first ionisation enthalpy) |
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Answer» The correct option is : (a) I < Br < CI < F (increasing electron gain enthalpy) (d) B < C < N < O (increasing first ionisation enthalpy) Explanation: The correct order of increasing negative electron gain enthalpy is : I < Br < F < Cl and the correct order of increasing first ionisation enthalpy is B < C < O < N. |
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| 1595. |
Which statement is wrong? (a) Bond energy of F2 > Cl2 (b) Electronegativity of F > Cl (c) F is more oxidising than Cl (d) Electron affinity of Cl > F |
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Answer» The correct option is : (a) Bond energy of F2 > Cl2 Explanation: Due to more repulsion in between non-bonding electron pair (2p) of two fluorines (due to small size of F-atom) in comparison to non-bonding electron pair (3p) in chlorine, the bond energy of F2 is less than Cl2. BE (F2) = 158.5 kJ/mole and BE (Cl2) = 242.6 kJ/mole. |
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| 1596. |
The ions O2- , F- , Na+, Mg2+ and Al3+ are isoelectronic. Their ionic radii show (a) a significant increase from O2- to Al3+ (b) a significant decrease from O2- to Al3+ (c) an increase from O2- to F- and then decrease from Na+ to Al3+ (d) a decrease from O2- to F- and then increase from Na+ to Al3+ |
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Answer» The correct option is : (b) a significant decrease from O2- to Al3+ Explanation: Amongst isoelectronic ions, ionic radii of anions is more than that of cations. Further size of the anion increases with increase in -ve charge and size of cation decreases with increase in +ve charge. Hence, correct order is O2- > F- > Na+ > Mg2+ > Al3+ |
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| 1597. |
With which of the following electronic configuration an atom has the lowest ionisation enthalpy? (a) 1s2 2s2 2p3 (b) 1s2 2s2 2p5 3s1 (c) 1s2 2s2 2p6 (d) 1s2 2s2 2p5 |
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Answer» The correct option is: (b) 1s2 2s2 2p5 3s1 Explanation: The larger the atomic size, smaller is the value of the ionisation enthalpy. Again higher the screening effect, lesser is the value of ionisation potential. Hence option (b) has lowest ionisation enthalpy. |
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| 1598. |
Specific volume of cylindrical virus particle is `6.02xx10^(-2) c c//g` whose radius and length `7 Å` and `10 Å` respectively. If `N_(A)=6.02xx10^(23)`, find molecular weight of virus:A. 1.54 kg/mol.B. `1.54 xx 10^4` kg/mol.C. `3.08 xx 10^4` kg/mol.D. `3.08 xx 10^3` kg/mol. |
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Answer» Correct Answer - A Sp. Vol (vol. of 1gm ) cylindrical virus particle `=6.02xx10^(-2)` cc/gm radius of virus r=7Å= `7xx10^(-8)` cm length of virus =`pir^2l` `=22/7xx(7xx10^(-8))^2xx10xx10^(-8)=154xx10^(-23)` cc wt. of one virus particle =`"Vol."/"Sp. vol."` `rArr (154xx10^(-23))/(6.02xx10^(-2))` gm `therefore` mol. wt. of virus =wt. of `N_A` particles `=(154xx10^(-23))/(6.02xx10^(-2))xx6.02xx10^(+23)` gm/mol =15400 gm/mol =15.4 kg/mol |
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| 1599. |
If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`A. `6.02 xx 10^(14) mol^(-1)`B. `6.02 xx 10^(15) mol^(-1)`C. `6.02 xx 10^(16) mol^(-1)`D. `6.02 xx 10^(17) mol^(-1)` |
| Answer» Correct Answer - D | |
| 1600. |
The correct of decreasing second ionisation enthalpy of `Ti(22),V(23),Cr(24)` and `Mn(25)` isA. `Mn gt Cr gt Ti gt V`B. `Ti gt V gt Cr gt Mn`C. `Cr gt Mn gt V gt Ti`D. `V gt Mn gt Cr gt Ti` |
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Answer» Correct Answer - C As ionization enthalpy (both first and second) increases from left to right across the period. Only chromium is exceptional due to the stable configuration `(3d^(5))` so the correct order is : `Crgt Mn gt V gt Ti` |
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